Question

A high-speed drill reaches $2000\text{rpm}$in$0.50\text{s}$ .

a. What is the drill's angular acceleration?

b. Through how many revolutions does it turn during this first $0.50\text{s}$?

Short answer

(a) The angular acceleration is$419\text{rad}/{\text{s}}^2$

(b) The number of revolution is $8.30\text{rev}$.

Step-by-step solution

Step 1. Given information

(a) The change in angular velocity,$\Delta \omega =2000\frac{\text{rev}}{\text{min}}$

(b) The change in time interval, $\Delta t=0.50\text{s}$

The expression for angular displacement in a given time interval is,

$\theta \left(t\right)={\omega }_i\Delta t+\frac{1}{2}\alpha {\left(\Delta t\right)}^2$

Here, ${\omega }_i$is the initial velocity, $\Delta t$is the time interval and is angular acceleration.

Part (a)

The angular acceleration.

$\begin{array}{l}\alpha =\frac{\Delta \omega }{\Delta t}\\ \alpha =\frac{2000\text{rpm}\left(\frac{2\pi }{1\text{rotation}}\right)\left(\frac{1\text{min}}{60\text{s}}\right)}{0.5\text{s}}=418.6\text{rad}/{\text{s}}^2\\ \alpha \approx 419\text{rad}/{\text{s}}^2\end{array}$

Part (b)

The number of revolutions turned during the first is given by,

$\begin{array}{l}\theta \left(t\right)={\omega }_i\Delta t+\frac{1}{2}\alpha {\left(\Delta t\right)}^2\\ \theta \left(t\right)=\left(0\right)\Delta t+\frac{1}{2}\left(419\text{rad}/{\text{s}}^2\right){\left(0.5\text{s}\right)}^2\\ \theta \left(t\right)=52.4\text{rad}\left(\frac{1\text{rev}}{2\pi \text{rad}}\right)=8.30\text{rev}\end{array}$

Therefore, the number of revolution$0.5\text{s}$ in is$8.30\text{rev}$ .